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zysi [14]
3 years ago
10

. A two-dimensional fluid motion is represented by a point vortex of strength Γ set at distance c from an infinite straight soli

d boundary. Write expressions for the velocity potential and stream function in Cartesian coordinates. Derive an expression for velocity on the boundary. Draw the streamlines and plot the velocity distribution on the boundary when Γ = π and c = 1.

Engineering
1 answer:
Lostsunrise [7]3 years ago
8 0

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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At the instant shown car A is travelling with a velocity of 24 m/s and which is decreasing at 4 m/s2 along the highway. At the s
SVEN [57.7K]

(a) V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

<u>Explanation:</u>

We will solve with respect to Cartesian vector form.

So,

V(A)= (24i) m/s

acc(A) = (4i) m/s²

There are two components of Car B, cos 60⁰ and sin 60⁰

V(B) = 20 cos 60° i + 20 sin 60° j

V(B) = (10 i + 17.32 j ) m/s

The car B moves along a curve, so it will have a tangential acceleration and a normal acceleration.

The tangential acceleration, a(t) = 5 m/s²

Normal acceleration, a(n) = \frac{v^2}{p} \\\\

So,

a(n) = \frac{(20)^2}{250}\\ \\a(n) = 1.6 m/s^2

For the tangential acceleration, the acceleration is slowing down. So,

a(t) = (-5 cos 60° i - 5 sin 60° j ) m/s²

a(t) = ( -2.5 i - 4.33 j) m/s²

For normal acceleration, it towards center. So,

a(n) = (1.6 sin 60° i - 1.6 cos 60° j) m/s²

a(n) = (1.39 i - 0.8 j ) m/s²

Total acceleration of Car B:

acc(B) = a(t) + a(n)

acc(B) = ( -2.5 i - 4.33 j) m/s² + (1.39 i - 0.8 j ) m/s²

acc(B) = (-1.11i - 5.13 j ) m/s²

(a) V(A/B) = ?

V(A) = V(B) + V(A/B)

(24i) m/s = (10 i + 17.32 j ) m/s + V(A/B)

V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ?

acc(A) = acc(B) + acc(A/B)

(4i) m/s² = (-1.11i - 5.13 j ) m/s² + acc(A/B)

acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

3 0
3 years ago
Draw a sinusoidal signal and illustrate how quantization and sampling is handled by
erica [24]
Answer: Your mums gay

Explanation: Your mums gay
3 0
3 years ago
A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu
trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the  first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

                              Area of  A is  = \frac{\pi}{4}(0.01)^{2} m^{2}

                                                    = 7.85 *10^{-5}m^{2}

Velocity of liquid through A = 0.2 m/s

  The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 0.2 *7.85*10^{-5} \frac{m^{3}}{s}

  The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              =  1039.8 * 0.2 * 7.85 *10^{-5} Kg/s

                              = 0.016324 \frac{Kg}{s}

From the question the diameter of B = 2 cm = 0.02 m

                                           Area of B = \frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}

                                     Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 3.14*10^{-4} *0.01 \frac{m^{3}}{s}

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 1053 * 3.14*10^{-6} \frac{Kg}{s}

                              = 0.00330642 \frac{Kg}{s}

From the question The flow rate in term of volume of the outflow at the time of measurement is given as  = 0.5 L/s

And also from the question the mass of  potassium chloride  at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 13\frac{g}{L} * 0.5 \frac{L}{s}

                              =  \frac{6.5}{1000}\frac{Kg}{s}       Note (1 Kg = 1000 g)

                              = 0.0065 kg/s

Considering potassium chloride

         Let denote the  rate at which liquid flows in terms of mass as   as \frac{dm}{dt} i.e change in mass with respect to time hence

           Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

         

      (0.016324 + 0.00330642) - 0.0065 = \frac{dm}{dt}

          \int\limits {\frac{dm}{dt} } \, dx  =\int\limits {0.01313122} \, dx

      => 0.01313122 t = (m - m_{o})

  From the question  (m - m_{o})  is given as = 19.7 Kg

Hence the time when the sample was taken is given as

               0.01313122 t = 19.7 Kg

      =>  t = 1500.2414 sec

            t = .4167337 hours (1 hour = 3600 seconds)

5 0
4 years ago
The locations and type of electrical device required on an architectural<br> plan are referred to as
Vikki [24]

The locations and type of electrical devices required on an architectural plan are referred to as <u>electrical symbols</u>. The use of detailed electrical plan ensures that all electrical equipment and wiring is installed exactly as intended. If the electrical plans are insufficient or hazy, the installation is left to the electrician’s discretion.  

Electrical plans are prepared using a CAD layered floor plan. Designers should not rely on electricians to design the electrical system, only to install it. Conversely, designers do not plan the position of every wire, only the position and relationship of all fixtures, devices, switches, and controls. This is done with the use of electrical symbols.  

Hundreds of electrical symbols are used on floor plans to describe what and where electrical elements will be installed. On simple plans, electrical symbols are often included as a separate layer on the floor plan. For larger or more complex structures, a separate plan is prepared.

The locations and type of electrical device required on an architectural plan are referred to as <u>electrical symbols</u>.

Learn more:

https://brainly.in/question/40648976

5 0
2 years ago
Read 2 more answers
Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a
stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0
3 years ago
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